2.2.3 Reaction Calculations

Mass Calculations

The number of moles of a substance can be found by using the following equation:

n u m b e r o f m o l e s (m o l) = \frac{m a s s o f a s u b s t a n c e (g)}{m o l a r m a s s (g m o l^{- 1})}

It is important to be clear about the type of particle you are referring to when dealing with moles
- E.g. one mole of CaF₂ contains one mole of CaF₂ formula units, but one mole of Ca²⁺and two moles of F^-ions

Reacting masses

The masses of reactants are useful to determine how much of the reactants exactly react with each other to prevent waste
To calculate the reacting masses, the balanced chemical equation is required
This equation shows the ratio of moles of all the reactants and products, also called the stoichiometry, of the equation
To find the mass of products formed in a reaction the following pieces of information are needed:
- The mass of the reactants
- The molar mass of the reactants
- The balanced equation

Worked Example

Mass calculation using moles

Calculate the mass of magnesium oxide that can be made by completely burning 6 g of magnesium in oxygen.

magnesium (s) + oxygen (g) → magnesium oxide (s)

Answer

Step 1: The balanced symbol equation is:

2Mg (s) + O₂ (g) → 2MgO (s)

Step 2: The relative formula masses are:

Mg = 24.3, O₂ = 32.0, MgO = 40.3

Step 3: Calculate the moles of magnesium used in the reaction:

n u m b e r o f m o l e s =

\frac{6.0 g}{24.3 g m o l^{- 1}}

= 0.25 moles

Step 4: Find the ratio of magnesium to magnesium oxide using the balanced chemical equation

Atoms, Molecules & Stoichiometry Table 1, downloadable AS & A Level Chemistry revision notes

Therefore, 0.25 mol of MgO is formed

Step 5: Find the mass of magnesium oxide

Mass = mol x M_r

Mass = 0.25 mol x 40 g mol^-1

Mass = 10 g

Therefore, the mass of magnesium oxide produced is 10 g

Stoichiometric relationships

The stoichiometry of a reaction can be found if the exact amounts of reactants and products formed are known
The amounts can be found by using the following equation:

n u m b e r o f m o l e s (m o l) = \frac{m a s s o f a s u b s t a n c e (g)}{m o l a r m a s s (g m o l^{- 1})}

The gas volumes can be used to deduce the stoichiometry of a reaction
- E.g. in the combustion of 50 cm³of propane reacting with 250 cm³of oxygen, 150 cm³ of carbon dioxide is formed suggesting that the ratio of propane : oxygen : carbon dioxide is 1 : 5 : 3

C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) + 4H₂O (l)

Volume Calculations

The concentration of a solution is the amount of solute dissolved in a solvent to make 1 dm³of solution
- The solute is the substance that dissolves in a solvent to form a solution
- The solvent is often water

$c o n c e n t r a t i o n (m o l d m^{- 3}) = \frac{n u m b e r o f m o l e s (m o l)}{v o l u m e o f s o l u t i o n (d m^{3})}$

A concentrated solution is a solution that has a high concentration of solute
A dilute solution is a solution with a low concentration of solute
When carrying out calculations involve concentrations in mol dm^-3the following points need to be considered:
- Change mass in grams to moles
- Change cm³to dm³
To calculate the mass of a substance present in solution of known concentration and volume:
- Rearrange the concentration equation

number of moles (mol) = concentration (mol dm^-3) x volume (dm³)

- Multiply the moles of solute by its molar mass

mass of solute (g) = number of moles (mol) x molar mass (g mol^-1)

Worked Example

Calculating volume from concentration

Calculate the volume of hydrochloric acid of concentration 1.0 mol dm^-3 that is required to react completely with 2.5 g of calcium carbonate

Answer

Step 1: Write the balanced symbol equation

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Step 2: Calculate the amount, in moles, of calcium carbonate that reacts

$n u m b e r o f m o l e s (C a C O_{3}) = \frac{2.5 g}{100 g m o l^{- 1}} = 0.025 m o l$

Step 3: Calculate the moles of hydrochloric acid required using the reaction’s stoichiometry

1 mol of CaCO₃ requires 2 mol of HCl

So 0.025 mol of CaCO₃ requires 0.05 mol of HCl

Step 4: Calculate the volume of HCl required

$v o l u m e o f H C l (d m^{3}) = \frac{a m o u n t (m o l)}{c o n c e n t r a t i o n (m o l d m^{- 3})} = \frac{0.05 m o l}{1.0 m o l d m^{- 3}} = 0.05 d m^{3}$

Volume of hydrochloric acid = 0.05 dm³

Worked Example

Neutralisation calculation

25.0 cm³ of 0.050 dm^–3sodium carbonate was completely neutralised by 20.0 cm³ of dilute hydrochloric acid.Calculate the concentration in mol dm^–³ of the hydrochloric acid.

Answer

Step 1: Write the balanced symbol equation

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Step 2: Calculate the amount, in moles, of sodium carbonate reacted by rearranging the equation for amount of substance (mol) and dividing the volume by 1000 to convert cm³ to dm³

amount (Na₂CO₃) = 0.025 dm³ x 0.050 mol dm^-3 = 0.00125 mol

Step 3: Calculate the moles of hydrochloric acid required using the reaction’s stoichiometry

1 mol of Na₂CO₃ reacts with 2 mol of HCl, so the molar ratio is 1 : 2

Therefore 0.00125 moles of Na₂CO₃ react with 0.00250 moles of HCl

Step 4: Calculate the concentration, in mol dm^-3, of hydrochloric acid

$c o n c e n t r a t i o n o f H C l (m o l d m^{- 3}) = \frac{a m o u n t (m o l)}{v o l u m e (d m^{3})} = \frac{0.00250}{0.0200} = 0.125 m o l d m^{- 3}$

Volumes of gases

Avogadro suggested that ‘equal volumes of gases contain the same number of molecules’ (also called Avogadro’s hypothesis)
At room temperature (20 degrees Celsius) and pressure (1 atm) one mole of any gas has a volume of 24.0 dm³
This molar gas volume of 24.0 dm³mol^-1can be used to find:
- The volume of a given mass or number of moles of gas:

volume of gas (dm³) = amount of gas (mol) x 24 dm³mol^-1

The mass or number of moles of a given volume of gas:

$a m o u n t o f g a s (m o l) = \frac{v o l u m e o f g a s (d m^{3})}{24.0 (d m^{3} m o l^{- 1})}$

Worked Example

Calculation volume of gas using excess & limiting reagents

Calculate the volume the following gases occupy:

Hydrogen (3 mol)
Carbon dioxide (0.25 mol)
Oxygen (5.4 mol)
Ammonia (0.02 mol)

Calculate the moles in the following volumes of gases:

Methane (225.6 dm³)
Carbon monoxide (7.2 dm³)
Sulfur dioxide (960 dm³)

Answer

Atoms, Molecules & Stoichiometry Table 2, downloadable AS & A Level Chemistry revision notes

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Revision Notes

2.2.3 Reaction Calculations

Mass Calculations

Reacting masses

Worked Example

Stoichiometric relationships

Volume Calculations

Worked Example

Worked Example

Volumes of gases

Worked Example

Author: Richard

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